[thingy]

Here's the situation:
Methyl benzoate reacts with aqueous alkali to give the benzoate anion and methanol. After acidification benzoic acid can be isolated from the reaction mixture.
A mixture of 0.50 mL (0.55 g) of methyl benzoate and 7.0 mL of 10% w/v aqueous KOH solution is heated at reflux for 20 minutes. The benzoate anion formed initially is converted into the corresponding acid by acidification after hydrolysis is complete.
After proper drying of the benzoic acid the average student would find the mass of their product is 0.35 g.

I am asked to find the amount of KOH before reaction in mol- how do I do this when I don't know the mass of the KOH?

[Dan]

I am asked to find the amount of KOH before reaction in mol- how do I do this when I don't know the mass of the KOH?

You have been given the volume and concentration of the KOH solution...

How are volume, concentration and amount (ie moles) related?

[thingy]

Thanx! I didn't realise what w/v was but I still get the wrong answer.

0.1 x 7 = 0.7 g
0.7[g] / 56.1[g/mol] = 0.0125 mol

What am I doing wrong?

[Dan]

What makes you think you're wrong?

[thingy]

What makes you think you're wrong?

This is for an online assignment. The program won't accept 0.0125 as the right answer.

[UG]

That's because you should use the masses they give you M(K = 39.10, C = 12.01, H = 1.01, O = 16.00) g mol^-1

Answer should be 1.2475495 x 10^-2 

[Borek]

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[UG]

Ya  sorry, bad habit